MCQ Questions for Class 11 Commerce Applied Mathematics Numerical Applications Quiz 7

Pipe A fills a tank of

$700$ litres capacity at the rate of

$40$ litres a minute. Another pipe B fills the same tank at the rate of

$30$ litres a minute. A pipe at the bottom of the tank drains the tank at the rate of

$20$ litres a minute. If pipe A is kept open for a minute and then closed and pipe B is kept open for a minute and then closed and then pipe C is kept open for a minute and then closed and the cycle repeated, how long will it take for the empty tank to overflow?

0%
$42$ minutes
0%
$30$ minutes
0%
$12$ minutes
0%
$4$ minutes

Explanation

In 3 Minutes Volume Supplied

$\Rightarrow 40L+30L-20L=50L$

So, In 1 Minute Volume Supplied will be

$\Rightarrow \dfrac{50L}{3}$

Time Taken to Fill The tank Completely

$\Rightarrow \dfrac{700L}{\dfrac{50L}{3}}=42mins$

Therefore Answer is

$42mins$

Two pipes A and B can fill a tank in

$36$ hours and

$45$ hours respectively. If both the pipes are opened simultaneously, how much time will be taken to fill the tank?

0%
$50$ hours
0%
$6$ hours
0%
$20$ hours
0%
$10$ hours

Explanation

Part filled by

$A$ in

$1$ hour

$=\dfrac{1}{36}$
Part filled by

$B$ in

$1$ hour

$=\dfrac1{45}$ ;
Part filled by

$(A + B)$ in

$1$ hour

$=\dfrac1{36}+\dfrac1{45}=\dfrac9{180}=\dfrac1{20}$
Hence, both the pipes together will fill the tank in

$20$ hours.

Two pipes A and B can fill a cistern in

$\displaystyle 37\frac{1}{2}$ min and

$45$ min respectively. Both pipes were opened and the cistern was filled in just

$\displaystyle \frac{1}{2}$ an hour if the pipe B is turned off after what time it is opened?

0%
$5$ min
0%
$9$ min
0%
$10$ min
0%
$15$ min

Explanation

Let B be turned off after x minutes. Then,

Part filled by (A + B) in x min. + Part filled by A in (30 -x) min. = 1.

$\therefore x\left(\dfrac{2}{75}+\dfrac{1}{45} \right)+(30-x)\dfrac{2}{75}=1$

$\Rightarrow \dfrac{11x}{225}+\dfrac{60-2x}{75}=1$

$\Rightarrow 11x+180-6x=225$

$\Rightarrow 5x=225-180$

$\Rightarrow 5x=45$

$\Rightarrow x=9$

A large tanker can be filled by two pipes

$A$ and

$B$ in

$60$ min and

$40$ min. respectively. How many minutes will it take to fill the tanker from empty state if

$B$ is used for half the time and

$A$ and

$B$ fill it together for the other half ?

0%
$15$ min
0%
$20$ min
0%
$27.5$ min
0%
$30$ min

Explanation

Part fill by (A+B) in 1 minute $=\dfrac{1}{60}+\dfrac{1}{40}=\dfrac{5}{120}=\dfrac{1}{24}$

Let the tank is filled in x minute then

$\dfrac{x}{2}\times \left(\dfrac{1}{24}+\dfrac{1}{40} \right)=1$

$\dfrac{x}{2}\times \dfrac{1}{15}=1$

$\dfrac{x}{30}=1$

$x=30 \ min$

A water tank can be filled by a tap in

$6$ hours while it can be emptied by another tap in

$8$ hours. If both the taps are opened simultaneously then after how much time will the tank get filled ?

0%
$14$ hours
0%
$32$ hours
0%
$18$ hours
0%
$24$ hours

Explanation

Let L be the capacity of tank in liters.

Tank will be filled at the rate

$\dfrac{L}{6}\,$ liters\hour.

While, The tank will get emptied at the rate

$\dfrac{L}{8}\,$ liters\hour.

$\therefore$ Effective rate of filling the tank is

$\dfrac{L}{6}-\dfrac{L}{8}=\dfrac{L}{24}$ liters/hour.

$\therefore$ The tank will get filled in

$L \times \dfrac{24}{L}=24$ hours.

Two pipes

$A$ and

$B$ together can fill a cistern in

$4$ hours. Had they been opened separately, then

$B$ would have taken

$6$ hours more than

$A$ to fill the cistern. How much time will be taken by

$A$ to fill the cistern separately ?

0%
$1$ hours
0%
$2$ hours
0%
$6$ hours
0%
$8$ hours

Explanation

Let the cistern be filled by pipe $A$ alone in $x$ hours

Then, pipe $B$ will fill it in $(x+6)$ hours.

$\therefore \displaystyle \frac{1}{x}+\frac{1}{x+6}=\frac{1}{4}$

$\Rightarrow \displaystyle \frac{x+6+x}{x(x+6)}=\frac{1}{4}$

$\Rightarrow x^2-2x-24=0$

$\Rightarrow x^2-6x+4x-24=0$

$\Rightarrow x(x-6)+4(x-6)=0$

$\Rightarrow (x-6)(x+4)=0$

If we neglect the negative value then

$x=6$

Hence the cistern be filled by pipe

$A$ alone in

$6$ hrs.

A tank is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the tank in the same time during which the tank is filled by the third pipe alone. The second pipe fills the tank five hour faster than the first pipe and 4 hour slower than the third. The time required by the first pipe is:

0%
$6$ hrs
0%
$10$ hrs
0%
$15$ hrs
0%
$30$ hrs

Explanation

Suppose first pipe alone takes $x$ hours to fill the tank

Then ,second and third pipe takes $(x-5)$ and $(x-9)$ hours to fill the tank.

$\therefore \displaystyle \frac{1}{x}+\frac{1}{x-5}=\frac{1}{x-9}$

$\Rightarrow \displaystyle \frac{x-5+x}{x(x-5)}=\frac{1}{x-9}$

$\Rightarrow (2x-5)(x-9)=x(x-5)$

$\Rightarrow x^2-18x+45=0$

$\Rightarrow x^2-15x-3x+45=0$

$\Rightarrow x(x-15)-3(x-15)=0$

$\Rightarrow (x-3)(x-15)=0$

$x=15 hrs.$

A booster pump can be used for filling as well as emptying a tank. The capacity of the tank is

$2400$

$\displaystyle m^{3}$ . The emptying capacity of the tank is 10

$\displaystyle m^{3}$ /min higher than its filling capacity and the pump needs 8 min lesser to empty the tank than it need to fill it. What is the filling capacity of the pump?

0%
$50$ $\displaystyle m^{3}/min.$
0%
$60$ $\displaystyle m^{3}/min.$
0%
$72$ $\displaystyle m^{3}/min.$
0%
None of these

Explanation

Let the filled capacity of the tank

$x \ m^3/minute$

Then emptied capacity of the tank

$=(x+10) \ m^3/minute$

$\therefore \dfrac{2400}{x}-\dfrac{2400}{x+10}=8$

$\Rightarrow 2400\left(\dfrac{10}{x(x+10)} \right)=8$

$\Rightarrow x(x+10)=3000$

$\Rightarrow x=50 \ m^3/min$

Pipes A and B can fill a tank in

$5$ and

$6$ hours respectively. Pipe C can empty it in

$12$ hours. If all the three pipes are opened together then the tank will be filled in

0%
$\displaystyle 1\frac{13}{17}$ hours
0%
$\displaystyle 2\frac{8}{11}$ hours
0%
$\displaystyle 3\frac{9}{17}$ hours
0%
$\displaystyle 4\frac{1}{2}$ hours

Explanation

Part filled by

$A$ in

$1$ hour

$=\dfrac{1}{5}$
Part filled by

$B$ in

$1$ hour

$=\dfrac1{6}$

Part emptied by

$C$ in

$1$ hour

$=\dfrac1{12}$ ;
Part filled by

$(A + B-C)$ in

$1$ hour

$=\dfrac1{5}+\dfrac1{6}-\dfrac1{12}=\dfrac{17}{60}$
Hence, three pipes together will fill the tank in

$3\dfrac{9}{17}$ hours.

$A$ takes twice as much time as

$B$ or thrice as much time as

$C$ to finish a piece of work. Working together, they can finish the work in

$2$ days.

$B$ can do the work alone in

0%
$3$ days
0%
$7$ days
0%
$4$ days
0%
$6$ days

Explanation

Suppose

$A, B$ and

$C$ take

$x\,,\,\dfrac{x}{2}\,\,$ and

$\dfrac{x}{3}$ days respectively to finish the work.
Then,

$\left (\dfrac{1}{x}+\dfrac{2}{x}+\dfrac{3}{x}\right)$

$=\dfrac{1}{2}$

$\therefore\,\,\dfrac{6}{x}=\dfrac{1}{2}$

$\therefore\,\,x=12$
So,

$B$ takes

$\dfrac{12}{2}=6$ days to finish the work.

Three pipes

$A, B$ and

$C$ can fill a tank in

$6$ hours. After working at it together for two hours,

$C$ is closed and

$A$ and

$B$ can fill the remaining part in

$7$ hours. The number of hours taken by

$C$ alone to fill the tank is :

0%
$10$
0%
$12$
0%
$14$
0%
$16$

Explanation

A,B and C fill tank in 6 hours. Then A,B and C one hours work = $\dfrac{1}{6}$

Part filled in two hours $=\dfrac{2}{6}$ = $\dfrac{1}{3}$

Then remaining part $=1-\dfrac{1}{2}=\dfrac{2}{3}$

A and B 7 hours work $=\dfrac{2}{3}$

Then A and B one hours work $=\dfrac{2}{21}$

So C one hours work $=(A+B+C)$ one hours work $-(A+B)$ one hours work $=\dfrac{1}{6}-\dfrac{2}{21}=$ $\dfrac{1}{14}$

So calone fill the tank in $14$ hours.

$x$ men consume

$x$ kg of good in

$x$ days, then

$y$ men will consume

$y$ kg of good in

0%
$y$ days
0%
$x$ days
0%
$\dfrac{x}{y}$ days
0%
$\dfrac{y}{x}$ days

Explanation

The number of men and number of days is in inverse proportion.
So constant of proportionality

$= k_1 = \cfrac {x}{y}$
Number of goods and number of days in direct proportion.
So constant of proportionality

$= k_2 = \cfrac {y}{x}$
So, number of days in which

$y$ mean will consume

$y$ kfg of food

$= k_1 \times k_2 \times x = \cfrac {x}{y} \times {y}{x} \times x = x$ days.

At what time are the hands of a clock together between 5 and 6 ?

0%
$33\displaystyle \frac{3}{11}$ min. past 5
0%
$28\displaystyle \frac{3}{11}$ Min. past 5
0%
$27\displaystyle \frac{3}{11}$ Min. past 5
0%
$26\displaystyle \frac{3}{11}$ Min. past 5

Explanation

If the time is 5 hours and

$x$ minutes after 5, when the two hands of the clock are same, then the angle formed by minute hand is:

$\cfrac{x}{60} \times 360 = 6x$
The hour hand moves

$\cfrac{1}{2}$ degree each minute, then angle traced by hour hand is =

$150 + \cfrac{x}{2}$
Hence,

$6x = 150 + \cfrac{x}{2}$

$150 = \cfrac{11x}{2}$

$x = \cfrac{300}{11}$

$x = 27 \cfrac{3}{11}$ min. past 5

If 40 persons consume 240 kg of rice in 15 days, in how many days will 30 persons consume 48 kg of rice?

0%
2 days
0%
3 days
0%
4 days
0%
5 days

Explanation

We first find out in which variations are the quantities related in the given situation.

When the number of people decreases from 40 to 30, the number of days to consume the rice will be more.
When the number of kg of rice decrease from 240 kg to 48 kg, number of days to consume will be less.

Hence, the number of days is in direct variation with the number of kg of rice and in inverse variation with the number of people.

Since number of days and number of kg of rice are in direct variation,
Constant of variation

$k_1 = \cfrac {48}{240}$
Since number of days and number of people are in inverse variation,
Constant of variation

$k_2 = \cfrac {40}{30}$
Now

$x = k_1 \times k_2 \times 15$

$= \cfrac {40}{30} \times \cfrac {48}{240} \times 15$

$= 4$
Hence,

$30$ people will consume

$4$ kg of rice in

$4$ days.

A can contains a mixture of two liquids

$A$ and

$B$ in the ratio

$7 : 5.$ When

$9$ litres of mixture are drawn off and the can is filled with

$B,$ the ratio

$A$ and

$B$ becomes

$7 : 9$ . How many litres of liquid

$A$ was contained by the can initially ?

0%
$10$
0%
$20$
0%
$21$
0%
$25$

Explanation

Suppose the can initially contains $7x$ and $5x$ litres of mixtures A and B respectively.

Quantity of A in mixture left $=(7x-\cfrac { 7 }{ 12 } \times 9)=(7x-\cfrac{21}{4})$ litres

Quantity of B in mixture left $=(5x-\cfrac { 5}{ 12 } \times 9)=(5x-\cfrac{15}{4})$ litres

Now,

$\cfrac{7x-\cfrac{21}{4}}{5x-\cfrac{15}{4}+9}=\cfrac{7}{9}$

$=>\cfrac{28x-21}{20x+21}=\cfrac{7}{9}$

$=>252x-189=140x+147$

$=>x=3$

Thus, the can contains $21$ litres of $A$ .

There is a pyramid on a base which is a regular hexagon of side 2a. If every slant edge of this pyramid is of length

$\displaystyle \frac{\left ( 5a \right )}{2}$ , then the volume of this pyramid must be

0%
$\displaystyle 3a^{3}$
0%
$\displaystyle 3a^{3}\sqrt{2}$
0%
$\displaystyle 3a^{3}\sqrt{3}$
0%
$\displaystyle 6a^{3}$

Explanation

Side of regular hexagon

$=2a$

$cm$

$\therefore$ ar. of hexagon

$=6\times \dfrac { \sqrt { 3 } }{ 4 } \times 2a\times 2a=6\sqrt { 3 } { a }^{ 2 }$

Slant height of pyramid

$=\dfrac { 5a }{ 2 } cm=AG$

$\therefore \quad h=\sqrt { { \left( \dfrac { 5a }{ 2 } \right) }^{ 2 }-{ \left( 2a \right) }^{ 2 } } =\dfrac { 3a }{ 2 }$

$\therefore$ Volume of pyramid

$=\dfrac { 1 }{ 3 } \times ar.base\times h$

$=\dfrac { 1 }{ 3 } \times 6\sqrt { 3 } { a }^{ 2 }\times \dfrac { 3a }{ 2 }$

$=3\sqrt { 3 } { a }^{ 3 }{ cm }^{ 3 }$

A is thrice as efficient as B and B is twice as efficient as C. if A, B and C work together , how long will they take to complete a job which B complete in

$10$ days ?

0%
$\displaystyle \frac{20}{9}$ days
0%
$\displaystyle \frac{11}{9}$ days
0%
$3$ days
0%
None of these

Explanation

Work efficiency ratio

$A:B=3:1\quad \times 2$

$B:C=2:1$

$A:B=6:2$

$B:C=2:1$

$A:B:C=6:2:1$

Time ratio

$A:B:C=1:2:6$

$1$ works is done by

$B$ in

$10$ days.

$2$ works is done by

$B$ in

$10$ days.

Sum of ratio=

$1+2+6=9$

$\therefore 1$ work is done by

$A,B$ and

$C$ =

$\dfrac{20}{9}$ days.

Find the day of the week on 18 July, 1776 (leap year).

0%
Monday
0%
Tuesday
0%
Wednesday
0%
Thursday

Explanation

Here

$1600$ years have '0' odd day.................. (A)

$100$ years have '5' odd days................. (B).

$75$ years = (18 leap years + 57 ordinary years)

$= (18 \times 2 + 57 \times 1)$

$= 93$ odd days

$= (7 \times 13 + 2) = 2$ odd days ....................(C)
Now, the number of days from 1st January to 18 July, 1776

$= 182 + 18 = 200$ days

$= (28 \times 7 + 4)$ days

$= 4$ odd days ....................(D)
Adding (A) + (B) +(C) +(D),
We get,

$0 + 5 + 2 + 4 = 4$ odd days
Hence, the day is Thursday.

A can do a job in 10 days and B in 20 days. They work together, but 4 days before the finish of the job A left. How many days they work together?

0%
$6$
0%
$\dfrac {16}3$
0%
$8$
0%
$14$

Explanation

Part of work done by

$A$ in

$1$ day=

$\dfrac{1}{10}$

Part of work done by

$B$ in

$1$ day=

$\dfrac{1}{20}$

$\because B$ alone worked during last

$4$ days.

$\therefore$ part of work done by

$B$ in

$4$ days=

$\dfrac{4}{20}=\dfrac{1}{5}$

part of work done by

$A$ and

$B$ (before

$4$ days of complete work) =

$1-\dfrac{1}{5}=\dfrac{5-1}{5}=\dfrac{4}{5}$

part of work done by

$A$ and

$B$ in

$1$ day=

$\dfrac{ 1 }{ 10 } +\dfrac{ 1 }{ 20 } =\dfrac{ 2+1 }{ 20 } =\dfrac{ 3 }{ 20 }$

$\dfrac{ 3 }{ 20 }$ part of work done by

$A$ and

$B$ in

$1$ day.

$1$ part of work done by

$A$ and

$B$ in

$\dfrac{1}{\dfrac{ 3 }{ 20 }}$ day =

$\dfrac{20}{3}$ days.

$\therefore \dfrac{4}{5}$ part of work is done by

$A$ and

$B$ in

$\dfrac{ 20 }{ 3 } \times \dfrac{ 4 }{ 5 }$ days

$=\dfrac{16}{3}$ days

$5\dfrac{1}{3}$ days.

Find the day of the week on 16 January, 1969

0%
Monday
0%
Tuesday
0%
Wednesday
0%
Thursday

Explanation

$1600$ years have '0' odd day..................... (A)

$300$ years have '1' odd day .......................(B)

$68$ years have

$17$ leap years and

$51$ ordinary years.
Thus

$(17 \times 2 + 51 \times 1)= 85$ odd days

$\displaystyle \cong$ '01' odd day.................... (C)
16 January has

$= 02$ odd days .........................(D)
Adding (A) + (B) +(C) +(D),
We get,

$0 + 01 + 01 +02 = 04$ odd days
Ans : Thursday

How many times are the hands of a clocks coincide in a day?

0%
$10$
0%
$11$
0%
$12$
0%
$22$

Explanation

The hands of the clock makes

$0$ degrees with each other

$11$ times in

$12$ hours.

Therefore, in

$24$ hours it would coincide

$22$ times.
Hence, option D is correct.

On what dates of October, 1975 did Tuesday fall ?

0%
$7^{th}, 14^{th}, 21^{st},28^{th}$
0%
$1^{st}, 8^{th}, 15^{th}, 22^{st},29^{th}$
0%
$2^{nd}, 9^{th}, 16^{st},23^{rd}, 30^{th}$
0%
$3^{rd}, 10^{th}, 17^{st}, 24^{th}, 31^{st}$

Explanation

For determining the dates, we find the day on 1st Oct, 1975.

$1600$ years have '0' odd days ............(A).

$300$ years have '01' odd days ............(B).

$74$ years have (18 leap years + 56 ordinary years)

$2 \times 18 + 1 \times 56 = 92$ odd days = '01' odd days.....(C)
Days from 1st January to 1st Oct., 1975
1st Jan - 30 June + 1st July to 1st Oct.

$181 + 31 + 31 + 30 + 1 = 274$ days $$

$$=\cfrac {274}{7} = '01'$$ odd days................... (D)

Adding (A) + (B) +(C) +(D)

$= 0 + 01 + 01 + 01$

$= '03'$ odd days
Ans. Wednesday( 1st Oct), hence

$7^{th}, 14^{th}, 21^{st},28^{th}$ Oct. will fall on Tuesday.

I carried

$1000$ kg of watermelon in summer by train. In the beginning, the water content was

$99\%$ . By the time I reached the destination, the water content had dropped to

$98\%$ . The reduction in the weight of the watermelon was

0%
$10 kg$
0%
$50 kg$
0%
$100 kg$
0%
$500 kg$

Explanation

Water + solid = 1000
Water is

$\displaystyle \frac{99}{100} \times 1000 = 990$
water evaporated is x.
so

$\displaystyle \left ( \frac{990 - x}{1000 - x} \right ) 100 = 98$

$99000 - 100 x = 96000 - 98 x$

$1000 = 2x \Rightarrow x = 500$

Find the day of the week on 26 January, 1950.

0%
Tuesday
0%
Friday
0%
Wednesday
0%
Thursday

Explanation

1600 years have '0' odd days ................ (1)
300 years have 1 odd day..........................(2)
49 years have 12 1eap years and 37 ordinary years.

$\displaystyle \therefore \quad \left( 12\times 2+37 \right) =\frac { 61 }{ 7 } =5$ odd years...................(3)
26 January has 5 odd days................... (4)
On adding (1), (2), (3) & (4),
11 odd days =04 odd days

$\displaystyle \therefore$ day is Thursday.

Find the day of the week on 15 August, 1947.

0%
Tuesday
0%
Friday
0%
Wednesday
0%
Thursday

Explanation

15th August 1947 has 1946 years and period from 1st January 1947 to 15th August 1947.
Now, first 1600 years have 0 days.
Next 300 years have 1 odd day.
46 years = 11 leap years + 35 non leap years =

$11\times 2 + 35 = 57$ days = 1 odd day
Number of days from 1st Jan 1947 to 15th August 1947 = 227 days = 3 odd days
Hence, total number of odd days =

$0 + 1 + 1 + 3 = 5$
Hence, if 0 odd day it is Sunday, 1 odd day it is a Monday and so on.
Hence 5 odd days is a Friday.

Find the day of the week on 26 January, 1950.

0%
Tuesday
0%
Friday
0%
Wednesday
0%
Thursday

Explanation

Total number of odd days in first 1600 yeas = 0
Number of odd days in next 300 years = 1
Number of odd days in 49 years =

$12 \times 2 + 37 \times 1 = 61$ days or 5 odd days
Thus total number of odd days in 1949 = 1 + 5 = 6 days
Number of days in 190 = 26 = 3 weeks + 5 days.
Thus, number of odd days in 1950 = 5

Hence, total number of odd days = 5 + 6 = 11 = 1 week + 4 days
Now, 0 odd days then Sunday, 1 odd day then Monday and so on
Since, 4 odd days hence it is Thursday.

If February 1, 1966 is Wednesday, what day is March 10, 1966 ?

0%
Monday
0%
Sunday
0%
Saturday
0%
Friday

Explanation

February 1, 1966 is a Wednesday. Number of days between 1st February and 10th March is =

$27 + 10 = 37$
Now,

$\cfrac{37}{7} = 5 + \dfrac{2}{7}$
Thus, there are 2 odd days between the two given dates. Hence, the 10th March will be: Friday (2 days after Wednesday)

Find the day of the week on 15 August, 1947.

0%
Tuesday
0%
Friday
0%
Wednesday
0%
Thursday

Explanation

$1600$ years have

$0$ odd days ...................(1)

$300$ years have

$1$ odd day......................... (2)

$46$ years have

$11$ leap years and

$35$ ordinary years.

$\displaystyle \therefore \quad \left( 11\times 2+35 \right) =57$ odd days =

$1$ odd day.
Till

$30$ June there are

$6$ odd days ............(3)
From

$1$ July to

$15$ August there are

$46$ days i.e.

$4$ odd days.

$\displaystyle \therefore$

$0 + 1 + 1 + 3 = 5$ odd days.

$\displaystyle \therefore$ The day will be Friday.

Find the day of the week on 26 January, 1950

0%
Tuesday
0%
Friday
0%
Wednesday
0%
Thursday

Explanation

Total number of odd days in first

$1600$ years

$= 0$
Number of odd days in next

$300$ years

$= 1$
Number of odd days in 49 years =

$12 \times 2 + 37 \times 1 = 61$ days or

$5$ odd days
Thus total number of odd days in year

$1949 = 1 + 5 = 6$ days
Number of days in year

$1950 = 26 = 3 weeks + 5$ days.
Thus, number of odd days in

$1950 = 5$

Hence, total number of odd days

$= 5 + 6 = 11 = 1$ week + 4 days
Now, 0 odd days then Sunday, 1 odd day then Monday and so on
Since, 4 odd days hence it is Thursday.

Twenty two men can complete a piece of work in

$17$ days. They work for

$2$ days. How many more men should now be employed so as to complete the work in another

$10$ days?

0%
$11$
0%
$22$
0%
$33$
0%
None of these

Explanation

Amount of work

$22$ men can do in

$1$ day

$= \dfrac {1}{17}$
So, Amount of work

$22$ men can do in

$2$ day

$= \dfrac {2}{17}$
Fraction of work left

$= \dfrac {12}{17=} \dfrac {15}{17}$
That is,

$1$ work done

$=22\times 17$
So,

$\dfrac {15}{17}$ work done

$=22 \times 17 \times \dfrac {15}{17}$

Men required to complete the work in another

$10$ days.
So, as per question, we have

$22\times 17\times \dfrac { 15 }{ 17 } =10\times x$

$\Rightarrow x=33$
So,

$11$ more men will be required.

How many times are the hands of a clocks coincide in a day ?

0%
$10$
0%
$11$
0%
$12$
0%
$22$

A fort had provision of food for

$150$ men for

$45$ days. After

$10$ days

$25$ men left the fort . The number of days for which the remaining food will last is__________.

0%
$\displaystyle 29\frac{1}{5}$
0%
$\displaystyle 37\frac{1}{4}$
0%
$42$
0%
$54$

Explanation

fort had provision of food for 150 men for 45 days,
After 10 days 25 men left the fort, so 125 men are there in the fort.
So, now fort is having food for 150 men for 35 days.

Suppose 125 men had food for x days.

Now, Less men, More days (Indirect Proportion)

$125\times x=150\times 35$

$x=42days$

Answer (C) 42days

There are taps

$A, B, C$ fill up at an independently in

$10 hr, 20 hr, 30 hr,$ respectively. Initially the tank is empty and exactly one pair of taps is open during each hour and every pair of taps is open at least for one hour, What is the minimum number of hours required to fill the tank?

0%
$8$
0%
$9$
0%
$10$
0%
$11$

Explanation

$A - 10 hr \quad B = 20 hr \quad C = 30 hr$
Exactly one pair of taps is open during each hour and every pair of taps is open at least for one hour. So, first we say,

$A$ and

$B$ are open for

$1$ hour, then

$B$ &

$C$ and then

$C$ &

$A$ .

$\displaystyle \left ( \frac {1} {10} + \frac {1} {20} \right) + \left ( \frac {1} {20} + \frac {1} {30} \right) + \left( \frac {1} {30} + \frac {1} {10} \right) = \frac {22} {60}$
First then second then Third then

In three hours the tank will be filled

$\left( \frac {22} {60} \right)^{th}$ part

Now, for minimum time, the rest tank must be filled with A and B taps.

$\displaystyle \left( \frac {1} {10} + \frac {1} {20} = \frac {9} {60} \right)$

So, the rest

$\displaystyle \left( \frac {38} {60} \right)^{th}$ Part of tank will take 5 hour more

So, the tank will be filled in

$8^{th}$ hour.

In U.P.on 17th Oct. 1996, the president rule was declared. Find the day of week on that date.

0%
Tuesday
0%
Friday
0%
Wednesday
0%
Thursday

Explanation

Formula to calculate day of week based on date given is

$=$ (Year Code + Month Code + Century Code + Number Date -Leap Year Code)

$mod \ 7$ .

Year code

$=$ (YY+(YY

$\div4)) \ mod \ 7$

$=(96+24) \ mod \ 7$

$=1$

Month code

$=0$

Century code

$=0$

Day code

$=(1+0+0+17) \ mod \ 7$

$=18 (mod \ 7)$

$=4$ .

Hence, the day is Thursday.

Find the day of the week on 15 August, 1947

0%
Tuesday
0%
Friday
0%
Wednesday
0%
Thursday

Explanation

$15$ th August

$1947$ has

$1946$ years and period from

$1$ st January

$1947$ to

$15$ th August

$1947$ .
Now, first

$1600$ years have 0 days.
Next

$300$ years have

$1$ odd day.

$46$ years =

$11$ leap years +

$35$ non leap years =

$11\times 2 + 35 = 57$ days =

$1$ odd day
Number of days from

$1$ st Jan

$1947$ to

$15$ th August

$1947$ =

$227$ days =

$3$ odd days
Hence, total number of odd days =

$0 + 1 + 1 + 3 = 5$
Hence, if

$0$ odd day it is Sunday,

$1$ odd day it is a Monday and so on.
Hence

$5$ odd days is a Friday.

If the day, two days after tomorrow be Thursday, what day would have been two days before yesterday ?

0%
Friday
0%
Tuesday
0%
Thursday
0%
Wednesday

A can do a piece of work in 7 days of 9 hr each and B can do it in 6 days of 7 hr each How long will they take to do it working together

$\displaystyle \frac{42}{5}$ hr a day?

0%
3 days
0%
4 days
0%
4.5 days
0%
None of these

Explanation

A can complete the work in

$7\times 9=63 hours$
B can complete the work in

$6\times 7=42 hours$
A's one hour work

$=\frac{1}{63}$
B's one hour work

$=\frac{1}{42}$
(A+B)'s one hour work

$=\frac{1}{63}+\frac{1}{42}=\frac{5}{126}$

$\therefore$ Both can finish the work in

$\frac{126}{5} hours$
Number of days of

$\frac{42}{5}$ hours each=

$\frac{126\times 5}{42\times 5}=3 days$

$5\,\text{January}\;1991$ was a Saturday, what day of the week was

$3\,\text{March}\;1992$ ?

0%
Saturday
0%
Sunday
0%
Monday
0%
Tuesday

Explanation

Analysis:

$1991$ is an ordinary year, and hence it has only one odd day. Thus,

$5\,\text{January}\;1992$ was one day after Saturday, i.e., Sunday.
Now, in January

$1992$ , there are

$26$ days remaining

$(31-5=26)$ , i.e.,

$5$ odd days

$(26\div7=3\,\text{weeks and 5 odd days})$ . In February

$1992$ , there are

$29$ days, i.e.,

$1$ odd day

$(29\div7=4\,\text{weeks and 1 odd day})$ .
In March

$1992$ , there are

$3$ odd days (we are calculating odd days only upto

$\text{3 March 1992}$ ).
Therefore, the total numbers of odd days after

$5\,\text{January}\;1992$ till

$3\,\text{March}\;1992$ is

$5+1+3=9 odd$ days, which is equivalent to

$2$ odd days

$(9\div 7= 1 week and 2 odd days)$ .
Therefore,

$\text{3rd March 1992}$ was

$2$ days after Sunday, i.e., Tuesday.

$A, B$ and

$C$ together can finish a piece of work in

$4$ days.

$A$ alone can do it in

$9$ days and

$B$ alone in

$18$ days. How many days will be taken by

$C$ to do it alone.

0%
$15$ days
0%
$14$ days
0%
$13$ days
0%
$12$ days

At what time are the hands of a clock together between 5 O'clock and 6 O'clock?

0%
$\displaystyle 33 \frac{3}{11}$ minutes past 5
0%
$\displaystyle 28 \frac{3}{11}$ minutes past 5
0%
$\displaystyle 27 \frac{3}{11}$ minutes past 5
0%
$\displaystyle 26 \frac{3}{11}$ minutes past 5

Explanation

If the time is 5 hours and

$x$ minutes after 5, when the two hands of the clock are same, then the angle formed by minute hand is:

$\cfrac{x}{60} \times 360 = 6x$
The hour hand moves

$\cfrac{1}{2}$ degree each minute, then angle traced by hour hand is =

$150 + \cfrac{x}{2}$
Hence,

$6x = 150 + \cfrac{x}{2}$

$150 = \cfrac{11x}{2}$

$x = \cfrac{300}{11}$

$x = 27 \cfrac{3}{11}$ min. past 5

Ramesh travels

$760km$ to his home partly by train and partly by car. He takes

$8hr$ , if he travels

$160km$ by train and the rest by car. He takes

$12$ minutes more, if he travels

$240km$ by train and the rest by car. Find the speed of train and the car.

0%
Speed of train $=80km/hr$ and speed of car $=100km/hr$
0%
Speed of train $=40km/hr$ and speed of car $=50km/hr$
0%
Speed of train $=60km/hr$ and speed of car $=120km/hr$
0%
Speed of train $=70km/hr$ and speed of car $=130km/hr$

Explanation

Let the speed of train be xkm/hr

and the speed car be ykm/hr respectively.

We know that

$Speed\quad =\dfrac { Distance }{ Time } \\ \Rightarrow Time\quad =\quad \dfrac { Distance }{ Speed }$

In Case 1

Distance travelled by train

$=$ 160km

Distance travelled by car

$=$ (760-160)km

$=$ 600km

Time taken by train+Time taken by car

$=$ 8 hours

Hence the equation becomes

$\dfrac { 160 }{ x } +\dfrac { 600 }{ y } \quad =\quad 8$

In case 2

Distance travelled by train

$=$ 240km

Distance travelled by car

$=$ (760-240)km = 520km

Time taken by train+Time taken by car

$=$ 8 hours+12 minutes

Hence the equation becomes

$\dfrac { 240 }{ x } +\dfrac { 520 }{ y } \quad =\quad \dfrac { 41 }{ 5 } ......(2)$

Hence we get two equations

$\dfrac { 160 }{ x } +\dfrac { 600 }{ y } \quad =\quad 8.........(1)\\ \quad \dfrac { 240 }{ x } +\dfrac { 520 }{ y } \quad =\quad \dfrac { 41 }{ 5 } ......(2)$

$Let\quad \dfrac { 1 }{ x\quad } =p\quad and\quad \dfrac { 1 }{ y } =q$

Hence we get equations

160p + 600q

$=$ 8........(3)

240p + 520q

$=$ 41/5...(4)

Solving equations (3) and (4)we get p

$=$ 1/80 and q

$=$ 1/100

$\quad Hence\quad p=\dfrac { 1 }{ x } \\ \Rightarrow x\quad =\dfrac { 1 }{ p } \quad =80\\ Similary\quad y\quad =\quad \dfrac { 1 }{ q } \quad =\quad 100$

Solving we get x

$=$ 80 and y

$=$ 100

hence, speed of train =80km/hr and speed of car =100km/hr.

On which day of the week was the New year's day celebrated in

$2007$ ?

0%
Sunday
0%
Monday
0%
Wednesday
0%
Friday
0%
Tuesday

If yesterday of yesterday was Wednesday, which of the following days would represent tomorrow's yesterday ?

0%
Tuesday
0%
Thursday
0%
Friday
0%
Saturday
0%
Wednesday

A clock is set right at

$\text{2 p.m.}$ If it gains one minute an hour, what is the true time when the clock indicates

$7$ pm the same day _______________.

0%
$55\displaystyle\frac{5}{61}\,\text{minutes past 6 pm}$
0%
$53\displaystyle\frac{5}{61}\,\text{minutes past 8 pm}$
0%
$52\displaystyle\frac{5}{61}\,\text{minutes past 4 pm}$
0%
$57\displaystyle\frac{5}{61}\,\text{minutes past 5 pm}$

At what time between

$\text{3 o ' clock and 4 o ' clock}$ , both the needles of the clock will coincide each other ?

0%
$16\displaystyle\frac{4}{11}\text{minutes past 3}$ .
0%
$\displaystyle\frac{2}{11}\,(3\times30+0)\,\text{minute past 3}$
0%
$\displaystyle\frac{3}{11}\,(3\times30+0)\,\text{minute past 3}$
0%
$\displaystyle\frac{4}{11}\,(3\times30+0)\,\text{minute past 3}$

$\text{26 January 1995}$ fell on a Thursday, on what day did it fall in the year

$1986$ ?

0%
Friday
0%
Monday
0%
Sunday
0%
Thursday
0%
Wednesday

$\text{8 March 1985}$ was Friday, then which was the day on

$\text{26 July 1985}$ ?

0%
Tuesday
0%
Friday
0%
Monday
0%
Thursday

A watch reads 4:If the minute hand points to East, in which direction does the hour hand point ?

0%
North-East
0%
South-East
0%
North-West
0%
North
0%
Blank

Which two months in a year have the same calender?

0%
April, July
0%
January, August
0%
March, October
0%
June, September

Anshu correctly remembers that his sister's b'day is before 20th but after 15th Oct. but her father correctly remembers that his daughter's b'day is before 17th Oct. Anshu's sister's b'day falls on which day?

0%
19th Oct
0%
16th Oct
0%
18th Oct
0%
15th Oct

CBSE Questions for Class 11 Commerce Applied Mathematics Numerical Applications Quiz 7 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 11 Commerce Applied Mathematics Numerical Applications Quiz 7 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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